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Question: is(-arctan(-x, y) = arctan(x, y))?

Question:is(-arctan(-x, y) = arctan(x, y))?

C_R 2632 Maple
arctan
May 07 2022
0

In the positive range Maple confirms that this is true.
In the real range Maple fails to provide an answer (see attachments).

Is this identity correct?

restart

kernelopts(version)

`Maple 2022.0, X86 64 WINDOWS, Mar 8 2022, Build ID 1599809`

 (1)

NULL

is(arctan(-x) = -arctan(x))

true

 (2)

`assuming`([is(arctan(x, y) = -arctan(-x, y))], [x::real, y::real])

FAIL

 (3)

`assuming`([is(arctan(x, y) = -arctan(-x, y))], [x::positive, y::positive])

true

 (4)

`assuming`([simplify(arctan(-x, y)+arctan(x, y), trig)], [x::real, y::real])

arctan(-x, y)+arctan(x, y)

 (5)

But

plot3d([arctan(-x, y)+arctan(x, y)], x = -1000000 .. 1000000, y = -1000000 .. 1000000, title = arctan(-x, y)+arctan(x, y))

  

On a unit circle

x = cos(alpha), y = sin(alpha)

x = cos(alpha), y = sin(alpha)

 (6)

subs(x = cos(alpha), y = sin(alpha), arctan(-x, y)+arctan(x, y))

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (7)

`assuming`([simplify(%)], [alpha::real])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (8)

`assuming`([simplify(%)], [alpha::positive])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (9)

`assuming`([simplify(%)], [-Pi < alpha and alpha < Pi])

arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha))

 (10)

plot(arctan(-cos(alpha), sin(alpha))+arctan(cos(alpha), sin(alpha)), alpha = -2*Pi .. 2*Pi, axes = boxed, color = red)

  

NULL


Download arctan_xy_simplify.mw

and another maybe related case where simplification does not work

arctan_xy_simplify_2.mw
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